Let $G$ a group and $H$ a subgroup (not necessarily normal), if we take the standard resolution $C_{\bullet}(G)$, we know that each $C_n(G)$ is a free (right) $\mathbb{Z}[G]$-module, but if we take $T$ as a set of repersentatives of cosets $G/H$ we have that $\mathbb{Z}[G]=\bigoplus_{g\in T}gH$ and then we have that $C_{\bullet}(G)$ is a free (right) $\mathbb{Z}[H]$-module, then the problem is passing an element of the homology group $H_n(C_{\bullet}(G)\otimes_{H}M)=H_n(H,M)$ to $H_n(C_{\bullet}(H)\otimes_{H}M)=H_n(H,M)$.
To do this, for example, A. Mazzoleni in his article "A New Proof of a Theorem of Suslin" (Page 205) uses this trick.
He takes a (set-theoretic) section $s:G/H\to G$ of the canonical projection $\pi:G\to G/H$ and defines a map of resolutions $C_{\bullet}(G)\to C_{\bullet}(H)$ taking every component $g_i$ and returning $\overline{g_i}:=s(\pi(g_i))^{-1}g_i$, i.e. $C_n(G)\to C_n(H)$ are defined by $$ (g_0,g_1,\cdots,g_n)\mapsto (\overline{g_0},\overline{g_1},\cdots,\overline{g_n}) $$
Making this, he obtain the same element in $H_n(H,M)$, in other words, the map is the identity in homology.
I need more documentation of this trick and how it works.
Thank You
